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I need to prove that this integral

$$A = \int \frac{e^{x^2}}{x} dx $$ is not solvable in elementary functions

What do I have as theory in my hands?

  1. I know 5 or 6 integrals for which it has been proven they are not solvable. One of them is this one

$$B = \int \frac{e^{x}}{x} dx $$

  1. I also know that if you transform an integral $A_1$ to $B_1$ with a substitution which involves just elementary functions, and if $B_1$ is not solvable, then $A_1$ is also not solvable.

These two results are stated in my real analysis book without proof. So I can just use them, this is my impression. I don't want to prove these two.

So to prove that $A$ is not solvable I do the substitution $x=\sqrt{t}$. This way I arrive at an integral which is $$\frac{1}{2} \cdot B$$

And since I know $B$ is not solvable, I conclude $A$ is also not solvable.

Is this a valid proof? I am not quite convinced if the substitution I did $x = \sqrt{t}$ is elementary.

peter.petrov
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  • It's pretty involved: https://math.stackexchange.com/questions/265780/how-to-determine-with-certainty-that-a-function-has-no-elementary-antiderivative – Chickenmancer Oct 16 '20 at 14:45
  • @Chickenmancer Thanks. Yeah... the general theory probably is. I don't want to go that far. I am just trying to solve an example using the theory I am given. – peter.petrov Oct 16 '20 at 14:48
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    Given your description of the context, this looks like a valid proof to me. Confirming that $\sqrt t$ is an elementary function should follow from whatever definition of elementary function you have on hand (and being able to do that verification is an important aspect of learning math). Depending on the definition, it might be helpful to note that $\sqrt t = \exp(\frac12\log t)$. – Greg Martin Oct 16 '20 at 16:03
  • @GregMartin Thanks, this was my thought too. That the power function $t^r$ where $r$ is real is also elementary (because it can be expressed as an elementary function of the exponential function), namely as you mentioned $\exp(r \cdot \ln{t})$. That argument should be correct, right? – peter.petrov Oct 16 '20 at 16:15

1 Answers1

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Here is a result that follows from Liouville's theorem (see https://math.stackexchange.com/a/3772327/589)

Let $f, g$ be rational functions, $g$ not constant. The indefinite integral $$ \int f(x)e^{g(x)}\;dx $$ is elementary if and only if there is a rational function $h$ such that $f=h'+hg'\;$.

Take $f(x) = 1/x$, $g(x) = x^2$. Is there a rational function $h$ such that $f=h'+hg'$, that is, $ 1 = xh'+2x^2 h$?

Write $h=p/q$ for coprime polynomials $p$ and $q$. Then $$ q^2=x(p'q-pq')+2x^2pq $$ and so $q$ divides $pq'$. Since $p$ and $q$ are coprime, $q$ divides $q'$ and so $q$ is constant. Therefore, $h$ is a polynomial. But then $1 = xh'+2x^2 h=x(h'+2xh)$ cannot happen for degree reasons.

lhf
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  • This is probably good but it's like teaching derivatives to somebody who just learnt to solve linear equations. It's well above my current level and purposes, I mean :) Thanks anyway. – peter.petrov Oct 16 '20 at 17:13